Secondlawproblems

Secondlawproblems - An inventor claims to lime dewluped u...

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Unformatted text preview: An inventor claims to lime dewluped u puwer e}ele capable of tleli\ering .1 net unrk output 014“) kl {or an energ)y input by heat transfer of llllltl k]. The \} stein undergoing the eyele reeehes the heat tmrtsl'er from hot games at a temperature 01' 500 K ii-yand discharges energy by heat trunxl'er ll) the atmosphere at 301) K E\illlellC thi\ elrtimt gsownom 2 _‘ Known: A s} stem nperutex in a eyele and produces a net amount «it vmrk while i'eeening .tntl dixehurging energy h) heat j transfer at fixed temperature» Find: EVuluute the claim that the eyele can develop 4H) kl of work for Lu] energ} input bx heat at 1000 kl. Schematic and Given Data: Qm = 1000 u Assumptions: e E_ The s) stem is 3hm~n on the Accompanying figure. 34 The hot gases and the .llllltlSpl‘lCl‘L‘ plu} the rules 01' hot and enld rexermrrs‘. respeethel). 4 £5993”? EELT Analysis: lnserting the \ulues supplied h) the inxentnr into Ed 52. the e}ele thermal el‘tieiene} l\ 411) U ,, 7] .~ "-— (l.~l| t-ll‘v'r) lllllll lxtl The maximum thermal el‘fieiene) (my ptmer e}ele eun haw \\ hile operating hemeen rexerwirx at 7H 7: 500 K and K 300 K ix giVen b} liq. 5.8 7; 300 K t 7)” I l - 4 l A ‘_~ ‘, '4 ”Alli-10W: " 7!; NM) k Since the thermal et'tieiene} «it the aetuul e}ele eweeds the muvmttm theoretical value the elumt eumint he \ztlid. WWW, .fi. , , ,,_.,,._‘W , W. ,WWMAW A,” ,~.,_,,...<,,... -W ,m _ 0 The temperatures used in e\ aluuttng 77mm mm] be in K or R M’Rfixfli Eta/@QLWQAM ’6”le EV‘CKWQM MW WmSgequLqemwxrcfi "Evy Y“ bk“ omn8\ S \AQxI‘P &.©, WW U53 0&3 . i" x A II M ‘35 5‘ fi; halllaflng mulhrfomanfife: 7.. ls of the freezer compartment. a refrigerator ’ it: .e or is at 22C. The rate of heat transfer hum-ff ..-J - nt at low temperature through passages in the wal t at ~ S'C when the air surrounding the refrigerat s 3000 um and the power input required to operate the refrigerator is 3200 km}; f the refrigerator and compare with the coefficient of performance of a reversible If; irs at the same two temperatures. , '3' _.4-. By xteadily circulating a refrigera titaintains the freezer oompanmen the freezer compartment to the refrigerant i Determine the ooet’rieient of performance 0 refrigeration cycle operating between reservo S O L U T i O N A refrigerator maintains a freezer compartment at a specified temperature. The rate of heat transfer from m, and the ambient temperature are known. - -‘_ Known: he power input to operate the refrigerator. refrigerated space, t Find: Determine the coefficient of performance and compare with that of a reverSiblc refrigerator operating between m4 voirs at the same two temperatures. - ' Schematic and Given Data: _= i . ) r: Assumptions: 2' ying figure is at steady 5m 2‘; _- "'t . l. The system shown on the accompan unding air play the roles off] 1. The freezer compartment and the surro and hot reservoirs. respectively. Freezer compartment zit—5°C (268 K) . 4 Figure 55.2 Inserting the given operating data into Eq. .. , B {QC , 8000ld/h 4 Wm“: 3200 kJ/h performance of a reversible refrigeration cycl Analysis: 2.5 Substituting values into Eq. 5.9 gives the coefficient of reservoirs at TC = 268 K and TH = 295 K TC 268 K ~' / : ~#~* ~ ~ 2 9.9 ‘3“ TH « TC 295 K — 268 K re may be sonic -.-_ . e suggests that the for - ._ . :1 nd maximum coefficients of performanc roached judiciously, however, rformance. This objective should be app implexity, and cost. ' _ ‘fiua‘kymeigk‘fi MCKMWWVK 400 J '- F‘%\<Z\QL\(L (gm/“Ligxmxsrwévkkk ”C2“ EL’NQSWV‘Q’QKA “if .n ‘ - ' V\§\gy\ oude 3kg»? (*5; ‘ XTQNALC‘AQAQ'. '7' i) The difference between the actual a for improving the thermodynamic pe performance may require increases in size. cr ...
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